dx/dy = (y+1)/x

x dx/dy = y + 1

Integrate both sides with respect to y,

∫ (x dx/dy) dy = ∫ (y + 1) dy

½x² = ½y² + y + c

Sub x = 1, y = 1,

½(1²) = ½(1²) + 1 + c

½ = 3/2 + c

c = -1

So,

½x² = ½y² + y - 1

x² = y² + 2y - 2

Complete the square.

x² = y² + 2y + 1 - 3

x² = (y + 1)² - 3

3 = (y + 1)² - x²

1 = (y + 1)²/3 - x²/3

Alternatively,


dx/dy = (y + 1)/x

dy/dx = x/(y + 1)

(y + 1) dy/dx = x

Integrate both sides with respect to x,

∫ ( (y + 1) dy/dx) dx = ∫ x dx

½y² + y = ½x² + c

Sub x = 1, y = 1,

½(1²) + 1 = ½(1²) + c

c = 3/2 - ½

c = 1

So,

½y² + y = ½x² + 1

y² + 2y = x² + 2

y² + 2y + 1 = x² + 3

(y + 1)² = x² + 3

(y + 1)² - x² = 3

(y + 1)² / 3 - x²/3 = 1

(Answer)

In your equation, you accidentally got the reciprocal of (y + 1) instead of (y + 1) itself.

Another possible presentation :

(y + 1) dy = x dx

∫ (y + 1) dy = ∫ x dx

½y² + y = ½x² + c

The rest of the working is the same as before.

when i move x to dx/dy side, does it not become 1/x?

No it doesn't.

You are basically multiplying both sides by x.

dx/dy = (y + 1)/x

x · dx/dy = x · (y + 1)/x

x · dx/dy = (y + 1)


If you choose to do the 'cross multiply way' ,


dx/dy = (y + 1)/x

x dx = (y + 1) dy

ohh ok, thanks for ur help :)

If you choose to do the 'cross multiply way'...

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It's not actually a "cross-multiplication" of the "dy", even though our brains often work that way for simplicity.

That's what the inverted commas were for.

The first two workings were presented in a way intended to lead the student towards the more proper way to present the working,

i.e reversal of the implicit differentiation

rather than 'cross multiplying' to get

① x dx = (y + 1) dy

Then adding the ∫ to get

② ∫ x dx = ∫ (y + 1) dy


She wrote her slightly erroneous version in this style perhaps due to the method she was coached or instructed.

The first step should be skipped and the second directly written down.

Presenting the first step would be technically wrong but if she was coached/taught to present it in that manner first before adding the second step, then by all means carry on.

So, that 'another possible presentation' was included for her to reference and follow in this respect.

In addition, most would simply write this directly :


dx/dy = (y + 1)/x

x dx/dy = y + 1

Integrate both sides with respect to y,

∫ x dx = ∫ (y + 1) dy


It wouldn't be absolutely necessary to show

∫ (x dx/dy) dy = ∫ (y + 1) dy

first.